(t+2)23+3(t+2)13−10=0 how do i solve?

Given equation is: ${\displaystyle \frac{{(t+2)}^2}{3}+3(t+2)\frac{1}{3}-10=0\dots \left(1\right)}$
Let ${\displaystyle \frac{{(t+2)}^1}{3}=x}$
${\displaystyle \frac{{(t+2)}^1}{3}(t+2)13+3(t+2)13-10=0}$
${\displaystyle x\times x+3\times x-10=0}$
${\displaystyle x^2+3x-10=0\dots \left(2\right)}$
Solving the quadratic equation (2) ${\displaystyle x^2+3x-10=0}$
${\displaystyle x^2+5x-2x-10}$
$x(x+5)-2(x+5)=0$
$(x+5)(x-2)=0$
$x=-5$ and $x=2$
Putting the value of x : ${\displaystyle \frac{{(t+2)}^1}{3}=x}$
${\displaystyle \frac{{(t+2)}^1}{3}=2}$
cubing both the sides ${\displaystyle (t+2)=2^3}$
$(t+2)=8$
$t=8-2=6$
$t=6$
$\frac{(t+2{)}^1}{3}=-5$
$(t+2)=-125$
$t=-125-2=-127$
Thus the value of t are 6 and -127